Every recurrence equation of the form
[tex]c_{n+1}=\frac{\kappa\,c_n+\lambda}{\mu\,c_n+\nu},\quad \kappa\,\nu-\mu\,\lambda\neq0[/tex]
can be brought into the linear form [itex]x_{n+1}=x_n+b_n[/itex] with [itex]b_n[/itex] known.
The transformation that does that, is
[tex]c_n=\frac{\alpha^{-n}}{x_n}+\beta[/tex]
Plug this to your equation and choose [itex]\alpha,\beta[/itex] in order to arrive to [itex]x_{n+1}=x_n+b_n[/itex]

Every recurrence equation of the form
[tex]c_{n+1}=\frac{\kappa\,c_n+\lambda}{\mu\,c_n+\nu},\quad \kappa\,\nu-\mu\,\lambda\neq0[/tex]
can be brought into the linear form [itex]x_{n+1}=x_n+b_n[/itex] with [itex]b_n[/itex] known.
The transformation that does that, is
[tex]c_n=\frac{\alpha^{-n}}{x_n}+\beta[/tex]
Plug this to your equation and choose [itex]\alpha,\beta[/itex] in order to arrive to [itex]x_{n+1}=x_n+b_n[/itex]

How do I get that to simplify into the required form? I tried plugging it in like this:

After making the substitution you suggested, I solved for [tex]\beta[/tex] by noticing that in order for the [tex]x_n x_{n+1}[/tex] term to disappear, either [tex]\alpha=0[/tex] or [tex]c_0^2 + 2c_0\beta -2\beta^2 = 0[/tex]. Solving, I got [tex]\beta=\pm \frac{c_0}{2} (1+\sqrt{3}).[/tex] For convenience, I only used [tex]\beta = \frac{c_0}{2} (1+\sqrt{3}).[/tex] Am I allowed to do that?

This is where I get "stuck". I figured out a way to solve for [tex]x_n[/tex] that's analagous to finding an integrating factor for a linear ODE, but it involves really nasty algebra. Is there a better way?